Question

hii.
Good morning....So today my question is in above pic....plz solve this...and plz give 10 thx to each of my questions i will follow u........​

Answer 1

Gɪᴠᴇɴ :-

$\bf\:x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a + 2b}-\sqrt{a-2b}}$

Tᴏ SHOW :-

• bx² - ax + b = 0

Sᴏʟᴜᴛɪᴏɴ :-

$\purple\longmapsto\tt\:x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a + 2b}-\sqrt{a-2b}}\\\\\textbf{Rationalizing The Denominator Now,}\\\\\purple\longmapsto\tt\:x=\frac{\sqrt{a + 2b} + \sqrt{a - 2b}}{ \sqrt{a + 2b} -\sqrt{a - 2b}} * \frac{\sqrt{a + 2b} +\sqrt{a - 2b}} {\sqrt{a + 2b} + \sqrt{a - 2b} }\\\\\purple\longmapsto\tt\:x= \frac{(\sqrt{a + 2b} + \sqrt{a - 2b})^2 }{(\sqrt{a + 2b}) ^2-(\sqrt{a + 2b}) ^2}\\\\ \purple\longmapsto\tt\:x=\frac{(\sqrt{a + 2b}) ^2+ ( \sqrt{a -2b}) ^2+2( \sqrt{a + 2b} )( \sqrt{a -2b} )}{a +2b - a + 2b}\\\\ \purple\longmapsto\tt\:x=\frac{a + 2b + a - 2b + 2( \sqrt{a + 2b})(\sqrt{a -2b} )}{4b} \\ \\\purple\longmapsto\tt\:x= \frac{2a + 2( \sqrt{a + 2b})(\sqrt{a -2b})}{4b} \\ \\\purple\longmapsto\tt\:x= \frac{2(a +\sqrt{a + 2b} \times\sqrt{a -2b}) }{4b} \\ \\\purple\longmapsto\tt\:x= \frac{(a +\sqrt{a + 2b} \times\sqrt{a -2b}) }{2b} \\ \\\textbf{Cross - Multiply Now,}\\\\\purple\longmapsto\tt\:2bx = a+ ( \sqrt{a + 2b}) ( \sqrt{a - 2b})\\\\\purple\longmapsto\tt \: 2bx - a = ( \sqrt{a + 2b}) ( \sqrt{a - 2b})$

Squaring Both Sides Now :-

⟼ (2bx - a)² = [√(a + 2b) * √(a - 2b)]²

⟼ 4b²x² + a² - 4bxa = ( a + 2b )( a - 2b )

⟼ 4b²x² + a² - 4bxa = a² - 4b²

⟼ 4b²x² + a² - 4bxa - a² + 4b² = 0

⟼ 4b²x² - 4bxa + 4b² = 0

⟼ 4b( bx² - xa + b ) = 0

⟼ bx² - ax + b = 0 (Hence, Proved).

$\rule{200}{4}$

Answer 2

GOOD MORNING MATE ❤️

Here's the solution⤴️